PHY
solutions_chapter19

# Solve a with the 1500 m resistor gives with the

• Notes
• 24
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 5–8. Sign up to view the full content.

Solve: (a) With the 1500 M resistor, gives With the resistor, gives Subtracting the second equation from the first gives and (b) Now R is 7.00 . Or, Reflect: The smaller R is, the larger the current and the smaller the terminal voltage. With the 1500 M resistor the current is very small and the terminal voltage differs only very little from the emf of the battery. 19.30. Set Up: When a battery is short-circuited, Solve: (a) (b) (c) r 5 6.00 V 1000 A 5 6.00 3 10 2 3 V 5 6.00 m V r 5 1.50 V 10.0 A 5 0.150 V r 5 E I 5 1.50 V 25.0 A 5 0.0600 V E 2 Ir 5 0. V V ab 5 IR 5 1.91 V. V ab 5 E 2 Ir 5 2.50 V 2 1 0.274 A 21 2.14 V 2 5 1.91 V. I 5 E R 1 r 5 2.50 V 7.00 V 1 2.14 V 5 0.274 A V E 5 1.75 V 1 1 0.350 A 2 r 5 2.50 V r 5 2.14 V . 0.75 V 5 1 0.350 A 2 r 1.75 V 5 E 2 1 0.350 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 1.75 V 5.00 V 5 0.350 A. 5.00 V 2.50 V 5 E 2 1 1.67 3 10 2 9 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 2.50 V 1500 3 10 6 V 5 1.67 3 10 2 9 A. V 1 E , r 2 V ab 5 IR . V ab 5 E 2 Ir . 11.3 V. IR 5 1 2.40 A 21 4.70 V 2 5 E 2 Ir 5 12.0 V 2 1 2.40 A 21 0.30 V 2 5 11.3 V. I 5 E r 1 R 5 12.0 V 4.70 V 1 0.30 V 5 2.40 A. E 2 Ir 2 IR 5 0 V 5 IR I 5 0, I 5 0, R 5 V I 5 2.97 V 1.65 A 5 1.80 V . V 2 IR 5 0 r 5 V 2 E I 5 3.08 V 2 2.97 V 1.65 A 5 0.067 V . I 5 1.65 A, V 5 2.97 V V 5 E 2 Ir , E 5 3.08 V. V 5 E 2 Ir . E . 15.0 A . I T 5 15.0 A 1 1 1 0.0020 1 2 2 1 21 100 C° 2 5 12.5 A. I 0 5 T 5 100°C T 5 0°C, I T 5 I 0 1 1 a 1 T 2 T 0 2 . I 0 5 V R 0 R T 5 R 0 3 1 1 a 1 T 2 T 0 24 . I new 5 V p d new 2 4 r L new 5 V p 1 2 d 2 2 4 r 1 2 L 2 5 2 I d new 5 2 d . L new 5 2 L ; Current, Resistance, and Direct-Current Circuits 19-5

This preview has intentionally blurred sections. Sign up to view the full version.

19.31. Set Up: When current passes through a battery in the direction from the terminal toward the terminal, the terminal voltage of the battery is Solve: (a) gives (b) so Reflect: The total resistance in the circuit is which agrees with the value specified in the problem. 19.32. Set Up: The graph shows when Solve: (a) When (b) When As A qualitative sketch of versus R is sketched in Figure 19.32a. Figure 19.32 (c) If then The graph of I versus R is sketched in Figure 19.32b. when and when is constant, independent of R. The graph of versus R is sketched in Figure 19.32c. 19.33. Set Up: For a resistor, and Solve: (a) (b) 19.34. Set Up: For a resistor, and Solve: (a) (b) 19.35. Set Up: Solve: (a) (b) 19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is Solve: (a) (b) Energy 5 Pt 5 1 300 W 21 3.0 3 10 2 3 s 2 5 0.90 J P 5 1 25 V 21 12 A 2 5 300 W. P 5 V ab I . Energy 5 1 1.17 W 21 1.5 h 21 3600 s / h 2 5 6320 J P 5 1 9.0 V 21 0.13 A 2 5 1.17 W Energy 5 Pt . P 5 VI . P 5 V 2 R 5 1 120 V 2 2 9.0 3 10 3 V 5 1.6 W V 5 IR 5 1 0.0183 A 21 15 3 10 3 V 2 5 275 V. I 5 Å P R 5 Å 5.0 W 15 3 10 3 V 5 0.0183 A. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI R 5 V I 5 15.0 V 21.8 A 5 0.688 V I 5 P V 5 327 W 15.0 V 5 21.8 A V 5 IR . P 5 VI V ab V ab V ab 5 E ; R S ` . I S 0 R S 0 I S ` I 5 E R . r 5 0 V ab R E (a) R I (b) V ab R E (c) V ab V ab S E . R S ` , V ab 5 0. R 5 0, V ab 5 IR 5 E R R 1 r 5 E 1 1 r / R . r 5 E I 5 5.00 V 2.5 A 5 2.00 V R 5 0, R 5 0. I 5 2.5 A V ab 5 E 2 Ir . I 5 E R 1 r . I 5 24.0 V 6.00 V 5 4.00 A, R 1 r 5 6.00 V . R 5 V ab I 5 21.2 V 4.00 A 5 5.30 V . V ab 2 IR 5 0 r 5 E 2 V ab I 5 24.0 V 2 21.2 V 4.00 A 5 0.700 V . V ab 5 E 2 Ir V ab 5 E 2 Ir . V ab 1 2 19-6 Chapter 19
19.37.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern